Determine a line such that all its points lie at equal distance to three non-parallel planes.

Question

I am supposed to determine the parametric equation of a line such that all it's points lie at equal distance to the three planes, $$x+2y+2z+3=0$$ $$x-2y+2z-1=0$$ $$2x+y+2z+1=0$$ So far I've been able to determine the point where all the planes intersect, as well as all the intersection lines between the planes individually. However, I am unable to get much further. I've tried determining a line that goes through the intersection point of the planes and another point with equal distance to the planes, but I just end up with a horrible system of equations. I'm thinking that there must be a much simpler way of going about solving this problem.

Answer 1

You did right the first step: if the planes have a common point the line shall pass through it.
However the planes do not need in general to have a common point.

The concept to apply is that, given two planes, the points equi-distant from them lie on one of the two planes bisecting the angles between the given planes. If these are parallel, then there is only one bi-secting plane (that in between) if they are distinct, or infinite if they are coincident .

Then, given two non-parallel planes $\pi_1=0, \quad \pi_2=0$ with unitary normal vectors $\bf n_1,\; \bf n_2$, the bisecting plane :
- belongs to the sheaf $\lambda \pi_1+ \mu \pi_2=0$
- has a unitary normal vector proportional to $\bf n_1+\bf n_2$ (external angle) or $\bf n_1-\bf n_2$ (internal angle).

Thus, having three planes,
- take two couples of them (e.g. $\pi_1,\,\pi_2$ and $\pi_2,\,\pi_3$)
- determine the four (or less, if you do not use homogeneous coordinates) bisecting planes $\pi_{1,2,a},\, \pi_{1,2,b}, \, \pi_{2,3,a},\, \pi_{2,3,b}$
- any line given by the crossing of two planes $\pi_{1,2,x}$ & $\pi_{2,3,y}$ will have $d_1=d_2\,\& \,d_2=d_3$.

In conclusion, for three non-parallel planes we have $4$ equi-distant lines. Less than that if some of the planes are parallel. The above when the distance is measured in absolute terms. If on each plane a direction of its normal is chosen as to measure the distance in algebraic ($\pm$) terms, then the line is unique (or does not exist).

To better visualize the whole situation, let's reduce the problem in 2D.
Given three non-parallel lines, thus a non-degenerate triangle made by them, the points that have the same absolute distance from the three lines are $4$: the $C_k$ shown in the sketch.

Coming to your particular case, the three planes are concurrent in one point: the system has only one solution $P=(1,-1,-1)$.
The unit normals to the planes are $$ \eqalign{ & {\bf n}_{\,1} = 1/3\left( {1,2,2} \right) \cr & {\bf n}_{\,2} = 1/3\left( {1, - 2,2} \right) \cr & {\bf n}_{\,3} = 1/3\left( {2,1,2} \right) \cr} $$

Four bisecting planes are $$ \eqalign{ & \pi _{1,2,a} = x + 0y + 2z + 1 = x + 2z + 1 = 0 \cr & \pi _{1,2,b} = 0x + 2y + 0z + 2 = y + 1 = 0 \cr & \pi _{2,3,a} = {3 \over 2}x - {1 \over 2}y + 2z + 0 = 3x - y + 4z = 0 \cr & \pi _{2,3,b} = - {1 \over 2}x - {3 \over 2}y + 0z - 1 = x + 3y + 2 = 0 \cr} $$ they are of course all passing through the point P.

Then starting and taking $\pi _{1,2,a} $ and $ \pi _{2,3,a} $, the cross product of their normals is $(2,2,-1)$.
Therefore a first line is $$ l_{\,1} :\;{{x - 1} \over { 2}} = {{y + 1} \over 2} = {{z + 1} \over { - 1}} = t $$ In fact, inserting its generic point $P_1(t)=(1+2t,-1+2t,-1-t)$ into the (normalized) equations of the three planes we get $4/3t(1,-1,1)$.

And you can check that you get analogue results with the other three lines obtained by the combination of $\pi _{1,2,x} $ and $ \pi _{2,3,y} $.

Answer 2

First, how to build the enveloping cone of a sphere.

Given a sphere $S\to \lVert p-p_c\rVert=r$ and a line $L\to p = p_0+\mu \vec v$ with $p = (x,y,z)^{\dagger}$ and $p_0$ the vertex, their intersection $S\cap L$ is obtained by solving for $\mu$

$$ \lVert p_0-p_c +\mu \vec v\rVert^2=r^2 $$

so

$$ \mu = \frac{-2(p_0-p_c)\cdot\vec v\pm\sqrt{4\left((p_0-p_c)\cdot\vec v\right)^2-4\left(\lVert p_0-p_c\rVert^2-r^2\right)\lVert\vec v\rVert^2}}{2\lVert\vec v\rVert^2} $$

but if tangency is required then

$$ \left((p_0-p_c)\cdot\vec v\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2\right)\lVert\vec v\rVert^2 = 0 $$

but $\vec v = \frac{1}{\mu}(p-p_0)$ so after substitution we have the cone equation

$$ \left((p_0-p_c)\cdot(p-p_0)\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2 \right)\lVert p-p_0\rVert^2=0 $$

Given three planes

$$ \begin{array}{rcl} \Pi_1\to (p-p_1)\cdot \vec n_1 = 0\\ \Pi_2\to (p-p_2)\cdot \vec n_2 = 0\\ \Pi_3\to (p-p_3)\cdot \vec n_3 = 0 \end{array} $$

with

$$ \begin{array}{rcl} p & =& (x,y,z)^{\dagger}\\ p_1 & = & (-3,0,0)^{\dagger}\\ p_2 & = & (1,0,0)^{\dagger}\\ p_3 & = & (0,-1,0)^{\dagger}\\ \vec n_1 & = &(1,2,2)^{\dagger}\\ \vec n_2 & = & (1,-2,2)^{\dagger}\\ \vec n_3 & = & (2,2,2)^{\dagger}\\ p_0 & = & (1,-1,-1)^{\dagger} \end{array} $$

with $\Pi_1\cap\Pi_2\cap\Pi_3= p_0$ a generic right cone tangent to the three planes and with vertex at $p_0$ can be written as

$$ C\to \left((p_0-p_c)\cdot(p-p_0)\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2 \right)\lVert p-p_0\rVert^2 = 0 $$

Now considering $C$ and $\Pi_1$ if they are tangent with $p_1^*$ as tangency point, we have

$$ \nabla C(p_1^*) = \lambda_1\vec n_1,\ \ (p_1^*-p_1)\cdot\vec n_1 = 0 $$

or

$$ \begin{array}{rcl} ((p_0-p_c)\cdot(p_k^*-p_0))(p_0-p_c)-\left(\lVert p_0-p_c\rVert^2-r^2 \right)(p_k^*-p_0)& = &\lambda_k\vec n_k\\ (p_k^*-p_k)\cdot \vec n_k & = & 0 \end{array} $$

giving four equations in the unknowns $x_k^*,y_k^*,z_k^*,\lambda_k$

Here the tangency points $p_k^*$ are located along an infinite line so

for each $k$ we have

$$ M_k(p_c,r)\cdot (x_k^*,y_k^*,z_k^*,\lambda_k)^{\dagger} = b_k $$

should be such that $\det(M_k) = f_k(p_c,r) = 0$

and the relationship given by $f_k(p_c,r) = 0$ furnish the equidistant lines in the form

$$ L_j\to p_c = p_0+\omega_j r $$

For the actual example we have

$$ \begin{array}{rcl} f_1(p_c.r) & = & (3r-3-x_c-2y_c-2z_c)(3r+3+x_c+2y_c+2z_c)\\ f_2(p_c,r) & = & (3 r+1 - x_c + 2 y_c - 2 z_c)(3 r-1 + x_c - 2 y_c + 2 z_c) \\ f_3(p_c,r) & = & (3 r-1 - 2 x_c - y_c - 2 z_c) (3 r+1 + 2 x_c + y_c + 2 z_c) \end{array} $$

arranging and solving for $p_c$ we get one

$$ \begin{array}{rcl} 3 r - x_c - 2 y_c - 2 z_c & = & 3\\ 3 r - x_c + 2 y_c - 2 z_c & = & -1\\ 3 r - 2 x_c - y_c - 2 z_c & = & 1 \end{array} $$

$$ p_c = p_0 + \frac{3}{2}(0,0,1)^{\dagger}r $$

and another one by solving

$$ \begin{array}{rcl} 3 r + x_c + 2 y_c + 2 z_c & = & -3\\ 3 r + x_c - 2 y_c + 2 z_c & = & 1\\ 3 r + 2 x_c + y_c + 2 z_c & = & -1 \end{array} $$

$$ p_c = p_0 + \frac{3}{2}(1,1,-\frac{1}{2})^{\dagger}r $$

and another

$$ \begin{array}{rcl} 3 r - x_c - 2 y_c - 2 z_c & = & 3\\ 3 r - x_c + 2 y_c - 2 z_c & = & -1\\ 3 r + 2 x_c + y_c + 2 z_c & = & -1 \end{array} $$

giving

$$ p_c = p_0 + (-6,0,\frac 92)^{\dagger}r $$

etc.

Attached a plot showing the three planes and the three cones associated to the three center lines. Obviously, those center lines are the sought lines,